What is the minimum thickness of thin film required for constructive interference in the reflected light through it ?
(Given, the refractive index of the film `=1.5,` wavelength of the lilght incident on the film `=600` nm.

Options

A. (a)100 nm
B. (b)300 nm
C. (c)50 nm
D. (d)200 nm

Answer

To find the minimum thickness of a thin film required for constructive interference in reflected light, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Condition for Constructive Interference**:
The condition for constructive interference in a thin film is given by the formula:
\[ 2nt = (m + \frac{1}{2})\lambda \]
where:
- \( n \) is the refractive index of the film,
- \( t \) is the thickness of the film,
- \( \lambda \) is the wavelength of the light in vacuum,
- \( m \) is an integer (0, 1, 2, ...).

2. **Identify the Given Values**:
- Refractive index \( n = 1.5 \)
- Wavelength of light \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)

3. **Set Up the Equation for Minimum Thickness**:
For minimum thickness, we use \( m = 0 \):
\[ 2nt = \frac{1}{2}\lambda \]

4. **Rearranging the Equation**:
Rearranging the equation to solve for \( t \):
\[ t = \frac{\lambda}{4n} \]

5. **Substituting the Values**:
Now, substitute the values of \( \lambda \) and \( n \):
\[ t = \frac{600 \times 10^{-9}}{4 \times 1.5} \]

6. **Calculating the Denominator**:
Calculate \( 4 \times 1.5 \):
\[ 4 \times 1.5 = 6 \]

7. **Final Calculation**:
Now substitute back to find \( t \):
\[ t = \frac{600 \times 10^{-9}}{6} = 100 \times 10^{-9} \, \text{m} \]
Thus, the minimum thickness \( t \) is:
\[ t = 100 \, \text{nm} \]

### Final Answer:
The minimum thickness of the thin film required for constructive interference in the reflected light is **100 nm**.

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